EduFarming
Your list · 0हिं

Type at least two letters. Press Esc to close.

↑↓ navigate↵ openTAB cycle filter

Inverse Trigonometric Functions · Question 62 of 120

What is cos^-1(-1/3) in terms of this α = sin^-1(1/3)?

Question 62

HardOriginal question

Let α=sin⁡−1(13)\alpha = \sin^{-1}\left(\frac{1}{3}\right).

What is cos⁡−1(−13)\cos^{-1}\left(-\frac{1}{3}\right) in terms of this α=sin⁡−1(13)\alpha = \sin^{-1}\left(\frac{1}{3}\right)?

Options for question 62

Show the answer and explanation
CorrectNot quiteAnswer D

Correct answer: D

Explanation

  1. cos⁡−1(13)=π2−sin⁡−1(13)=π2−α\cos^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{2} - \sin^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{2} - \alpha.

  2. cos⁡−1(−13)=π−(π2−α)=π2+α\cos^{-1}\left(-\frac{1}{3}\right) = \pi - \left(\frac{\pi}{2} - \alpha\right) = \frac{\pi}{2} + \alpha.

Worth noting

π−α\pi - \alpha applies the rule of cos⁡−1\cos^{-1} to α\alpha, which is a sin⁡−1\sin^{-1} value; first write cos⁡−1(13)\cos^{-1}\left(\frac{1}{3}\right) as π2−α\frac{\pi}{2} - \alpha.

Concept
Principal values of related inputs
Typical time
150 s
Your time
—
The rule this testsPrincipal values
Practise this chapter →

Ten questions from Inverse Trigonometric Functions, at your own pace.

Practise this chapter →

Ten questions from Inverse Trigonometric Functions, at your own pace.

Practise this chapterAll 120 questions in Inverse Trigonometric FunctionsRevise the notes for Inverse Trigonometric Functions